Need help

[crude fern]

Try to formulate equations based on the given information. You didn't provide your workout, so I'm not sure where you're stuck.

[summer yarrow]

I don’t even know how to do it

[summer yarrow]

If I can use that way to work out the answer then I won’t even post it here

[crude fern]

a) Consider the number of gold balls as g. Now what is the probability of picking two gold balls randomly?
It is $\frac{g}{40} * \frac{g-1}{39}$.

It's given that the probability is $\frac{5}{12}$. So I think you can follow it from here.

[dapper stirrup]

Bocchi

[crude fern]

b) You already should know that geometric sequences can be represented as:
$a, ar, ar^2, ..., ar^{(n-1)}$
Now based on the given conditions, try to form two equations.
The first one states that the product of first and last term equals 3, so $a^2r^{n-1}=3$.
Follow the similar approach and write another equation.

[dapper stirrup]

Bocchi

[summer yarrow]

Why u have to times 39?

[crude fern]

According to our hypothesis we have already picked a gold ball, so the number of remaining gold balls is g-1, and the number of total remaining balls is 40-1 = 39.

[summer yarrow]

Ok

[summer yarrow]

Second one is about the sum of the sequence?

[crude fern]

It's about product, so you need to figure out the equation by multiplying n terms of the sequence.

[summer yarrow]

Ok

[summer yarrow]

a^nxr^(n-1)=59049?

[crude fern]

I don't think so, try to think about the r^(n-1) part again.

[summer yarrow]

Just n?

Like the term before that

[crude fern]

No, it's going to be $(r^{(n-1)})^{\frac{n}{2}}$ if n is even.

You have n terms, and taking first and last term, and multiplying them gives $a^2r^{n-1}$

You can make $n/2$ amount of pairs if n is even, and if n is odd then you need to consider about the middle term(which you can figure out).

For n being odd and even, solve these two cases separately.

[dapper stirrup]

Bocchi

[summer yarrow]

How does it do with odd or even?

[summer yarrow]

Is A) 26?

[crude fern]

Yes, looks correct.

[summer yarrow]

What?

It’s not 100% correct?

[crude fern]

You don't need to consider even, or odd terms specifically, I thought there would be a necessity, but there isn't. You can safely go with n(n-1)/2

[crude fern]

In my opinion, it is correct. But I am a human, so I prefer not being 100% confident on my answers.

[summer yarrow]

Ok?

[summer yarrow]

That is for addition?

[crude fern]

$a^b \times a^c = a^{b+c}$, That's why multiplying terms would lead to adding exponents. Since there will be n-1 terms (From 1 to n-1), the summation will be $\frac{n(n-1)}{2}$

[dapper stirrup]

Bocchi

[summer yarrow]

But that is not for geometric sequences

[crude fern]

According to the question's second condition, you're multiplying the terms. So while multiplying those terms, their exponents will keep get adding.

[summer yarrow]

T^1+…n?

[summer yarrow]

Is this correct?

[crude fern]

Looks correct to me.

[summer yarrow]

Suddenly being so smart at night for no reason. Whatever thx

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[crude fern]

Write .close

[summer yarrow]

.close

[high crag]

Solved

Post marked as solved by @summer yarrow.

Use .unsolved if this was a mistake.